Integral of Lorentzian type with trigonometric function

Question

Consider the following Riemann integral $$ \int_0^\infty \mathrm{d}x \frac{\alpha^2}{(x-x_0)^2+\alpha^2} \frac{\sin\left[{\left(x - x_1\right) t }\right]}{x-x_1} $$ with the displacements $x_0,x_1 \in \mathbb{R}$ and broadening $\alpha \in \mathbb{R}$. The (claimed) solution should be $$ \pi \left[\frac{\alpha^2}{\delta^2 + \alpha^2} + e^{-\alpha t}\frac{\delta^2 \sin{(\delta t)}- \alpha^2\cos{(\delta t)}}{\delta^2 + \alpha^2} \frac{}{} \right] $$ with $\delta = x_0 - x_1$. Is an approximation involved, and how do I evaluate this integral?

Answer 1

Start from a simplified form involving fewer constants, to grasp the idea; we need $$ \int_{-\infty}^{+\infty} \left(\frac{1}{1+t^2}\right) \frac{\sin t}{t}dt; $$ notice that the integrand is the product of two Fourier transforms: $$ \int_{-\infty}^{+\infty}\frac{1}{2}\chi_{[-1,1]}(x)e^{-itx}dx=\frac{\sin t}{t} $$ where $\chi$ denotes the characteristic function, and $$ \int_{-\infty}^{+\infty}\frac{1}{2}e^{-|x|}e^{-ixt}dx= \Re\int_{0}^{+\infty}e^{-x(1+it)}dx= \Re \frac{1}{1+it}=\frac{1}{1+t^2}. $$ Now, the $L^2$ product of the Fourier transforms is $2\pi$ times the $L^2$ product of the original functions, i.e.: $$ \int_{-\infty}^{+\infty} \left(\frac{1}{1+t^2}\right) \frac{\sin t}{t}dt= 2\pi\int_{-\infty}^{+\infty}\frac{1}{2}\chi_{[-1,1]}(x)\frac{1}{2}e^{-|x|}dx={\pi}\int_{0}^1e^{-x}dx={\pi(1-e^{-1})}. $$ Now, we try to put the constants in this framework.

Explicitly: for $x=x_0+s$, denoting $\delta\equiv x_0-x_1$, we get $$ \int_{-\infty}^{+\infty}\frac{\alpha^2}{s^2+\alpha^2}\frac{\sin((\delta+s)t)}{\delta+s}ds; $$ now, $$ \int_{-\infty}^{+\infty}\frac{\alpha e^{-\alpha|v|}}{2}e^{-iyv}dv=\frac{\alpha^2}{\alpha^2+y^2} $$ whereas $$ \int_{-\infty}^{+\infty}e^{-ivy}e^{-iv\delta }\frac{1}{2}\chi_{[-1,1]}\left(v/t\right)dv=\frac{\sin((\delta+y)t)}{\delta+y}, $$ by the above mentioned theorem: $$ \int_{-\infty}^{+\infty}\frac{\alpha^2}{s^2+\alpha^2}\frac{\sin((\delta+s)t)}{\delta+s}ds= 2\pi \int_{-\infty}^{+\infty} \frac{\alpha e^{-\alpha|v|}}{2} e^{-iv\delta }\frac{1}{2}\chi_{[-1,1]}\left(v/t\right) dv. $$ This integral is a bit lengthy, but easy to compute explicitly, yielding precisely: $$ \int_{-\infty}^{+\infty}\frac{\alpha^2}{s^2+\alpha^2}\frac{\sin((\delta+s)t)}{\delta+s}ds= \pi \left[ \frac{\alpha^2}{\alpha^2+\delta^2}+e^{-\alpha t}\frac{\alpha\delta \sin(\delta t) - \alpha^2 \cos(\delta t)}{\alpha^2+\delta^2} \right]. $$ [Side note: since the integrand falls off as $1/x^3$, the Lebesgue integral works just as fine in this case.]

Answer 2

I'm assuming the integral is from $-\infty$ to $ + \infty$.

If $\alpha, t >0$, we can integrate the complex function $$ f(z) = \frac{\alpha^{2}}{(z-x_{0})^{2}+\alpha^{2}} \frac{e^{izt}e^{-ix_{1}t}}{z-x_{1}}$$ around a closed semicircle in the upper half-plane that is indented at $z=x_{1}$.

Doing so, we get $$ \text{PV} \int_{-\infty}^{\infty} f(x) \ dx - i \pi \text{Res}[f(z), x_{1}] =2 \pi i \, \text{Res}[f(z), x_{0}+i \alpha], $$

where $$\text{Res}[f(z), x_{1}] = \frac{\alpha^{2}}{(x_{1}-x_{0})^{2}+\alpha^{2}} $$

and $$ \begin{align} \text{Res}[f(z), x_{0}+ i \alpha] &= \frac{\alpha^{2}}{2i \alpha} \frac{e^{i(x_{0}+i\alpha)t}e^{-i x_{1}t}}{x_{0}+i\alpha-x_{1}} \\&= \frac{\alpha}{2i} \frac{e^{i(x_{0}+i\alpha)t}e^{-i x_{1}t}(x_{0}-x_{1}-i\alpha)}{(x_{0}-x_{1})^{2}+\alpha^{2}} \\ &= \frac{\alpha}{2i} \frac{e^{-\alpha t}e^{i(x_{0}-x_{1})t}(x_{0}-x_{1}-i\alpha)}{(x_{0}-x_{1})^{2}+\alpha^{2}}. \end{align} $$

Equating the imaginary parts on both sides of the equation and then dropping the PV sign, we get

$$ \begin{align} &\int_{-\infty}^{\infty} \frac{\alpha^{2}}{(x-x_{0})^{2}+\alpha^{2}} \frac{\sin[(x-x_{1})t]}{x-x_{1}} \, dx \\ &= \frac{ \pi \alpha^{2}}{(x_{0}-x_{1})^{2}+\alpha^{2}} + \pi \alpha e^{-\alpha t} \, \frac{(x_{0}-x_{1})\sin[(x_{0}-x_{1})t]- \alpha \cos[(x_{0}-x_{1})t]}{(x_{0}-x_{1})^{2}+\alpha^{2}} , \end{align} $$ which is slightly different than the result you posted but agrees with Brightsun's answer.